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question:Calculate the result of (sqrt{2}sqrt{3})^{2020} cdot (sqrt{2}+sqrt{3})^{2021}. A: sqrt{2}+sqrt{3} B: sqrt{3}sqrt{2} C: sqrt{3}sqrt{2} D: sqrt{2}sqrt{3}
answer:To solve the given problem, we will break down the calculation into detailed steps: 1. **Expression Simplification**: - We start with the given expression (sqrt{2}-sqrt{3})^{2020} cdot (sqrt{2}+sqrt{3})^{2021}. - Notice that we can factor this expression as [(sqrt{2}-sqrt{3})(sqrt{2}+sqrt{3})]^{2020} cdot (sqrt{2}+sqrt{3}). 2. **Applying Difference of Squares**: - The term (sqrt{2}-sqrt{3})(sqrt{2}+sqrt{3}) is a classic example of the difference of squares formula, a^2 - b^2 = (a-b)(a+b). - Substituting a = sqrt{2} and b = sqrt{3}, we get (sqrt{2})^2 - (sqrt{3})^2 = 2 - 3. 3. **Simplifying the Power**: - The expression simplifies to (2-3)^{2020} cdot (sqrt{2}+sqrt{3}). - Since 2-3 = -1, we have (-1)^{2020} cdot (sqrt{2}+sqrt{3}). - The term (-1)^{2020} equals 1 because any even power of -1 is 1. 4. **Final Simplification**: - Therefore, the expression simplifies to 1 cdot (sqrt{2}+sqrt{3}). - This simplifies further to sqrt{2}+sqrt{3}. Thus, the final answer to the problem is boxed{A}, which corresponds to sqrt{2}+sqrt{3}.
question:The number of integer solutions to the equation ((x+1)^{2}+(y2)^{2}=1) is: (A) 1 (B) 2 (C) 4 (D) Infinite (From Beijing Middle School Mathematics Competition, 1990)
answer:1. Analyze the given equation ((x+1)^2 + (y-2)^2 = 1). 2. Recognize that this represents a circle centered at ((-1, 2)) with a radius of 1. 3. Determine the possible integer values for (x+1) and (y-2) since the equation must hold for integer solutions: - Since ((x+1)^2 geq 0) and ((y-2)^2 geq 0), the values for ((x+1)) and ((y-2)) must be between (-1) and (1). 4. The only possible integer values for ((x+1)) and ((y-2)) that maintain the equation ((x+1)^2 + (y-2)^2 = 1) are: - (x+1 = 0) and (y-2 = pm 1) - (x+1 = pm 1) and (y-2 = 0) 5. Evaluate each possibility: - (x+1 = 0 implies x = -1) - (y-2 = 1 implies y = 3) - (y-2 = -1 implies y = 1) - (x+1 = 1 implies x = 0) - (y-2 = 0 implies y = 2) - (x+1 = -1 implies x = -2) - (y-2 = 0 implies y = 2) 6. Therefore, the integer solutions are ((-1, 3)), ((-1, 1)), ( (0, 2) ), and ((-2, 2)). 7. Hence, there are a total of 4 integer solutions. Conclusion: [ boxed{4} ]
question:A hollow glass sphere with uniform wall thickness and an outer diameter of 16 mathrm{~cm} floats in water in such a way that frac{3}{8} of its surface remains dry. What is the wall thickness, given that the specific gravity of the glass is s = 2.523?
answer:1. **Determine the partially submerged height of the sphere**: Given that the sphere's outer diameter is 16 , text{cm} and frac{3}{8} of its surface remains dry, the submerged fraction is: [ frac{5}{8} text{ of the diameter} ] Therefore, the height of the submerged part ( m ) is: [ m = frac{5}{8} times 16 , text{cm} = 10 , text{cm} ] 2. **Calculate the volume of the glass spherical shell**: Let ( x ) be the inner radius of the glass sphere. The outer radius is ( 8 , text{cm} ). The volume ( V ) of the spherical shell is the difference between the volumes of two spheres: one with radius ( 8 , text{cm} ) and one with radius ( x ). [ V_{text{shell}} = frac{4}{3} pi (8^3 - x^3) = frac{4 pi}{3} (512 - x^3) ] 3. **Calculate the volume of the submerged spherical cap**: The volume ( V_{text{cap}} ) of a spherical cap can be computed from the formula: [ V_{text{cap}} = frac{1}{3} pi m^2 (3R - m) ] Substituting ( m = 10 , text{cm} ) and ( R = 8 , text{cm} ): [ V_{text{cap}} = frac{1}{3} pi (10)^2 left(3 times 8 - 10right) = frac{1}{3} pi (100)(24 - 10) = frac{1}{3} pi (100)(14) = frac{1400 pi}{3} text{ cm}^3 ] 4. **Apply Archimedes' principle**: According to Archimedes' principle, the buoyant force equals the weight of the displaced water, which means the weight of the submerged glass sphere must equal the weight of the water displaced by the submerged part. [ text{Weight of glass sphere} = text{Weight of displaced water} ] The weight of the glass sphere (density ( s = 2.523 )) is: [ (text{Volume of glass}) times s = left( frac{4 pi}{3} (512 - x^3) right) times 2.523 ] The weight of the displaced water (assuming water has a density of 1 (text{g/cm}^3)) is: [ frac{1400 pi}{3} text{ grams} ] Setting these equal gives: [ frac{4 pi}{3} (512 - x^3) times 2.523 = frac{1400 pi}{3} ] 5. **Solve for ( x )**: Cancel out common terms ((frac{4 pi}{3})) from both sides of the equation: [ (512 - x^3) times 2.523 = 350 ] Solving for ( x^3 ): [ 512 - x^3 = frac{350}{2.523} ] Simplify the right-hand side: [ x^3 = 512 - frac{350}{2.523} approx 512 - 138.7 approx 373.3 ] Taking the cube root of both sides: [ x approx sqrt[3]{373.3} approx 7.2 , text{cm} ] 6. **Determine the thickness of the glass**: The glass thickness is the difference between the outer radius and inner radius: [ text{Thickness} = 8 , text{cm} - 7.2 , text{cm} = 0.8 , text{cm} ] Conclusion: (boxed{0.8 , text{cm}})
question:Compute: frac {5}{ sqrt {2}} sqrt { frac {1}{2}} = _______ .
answer:Solution: The original expression equals frac {5 sqrt {2}}{2} - frac { sqrt {2}}{2} = 2 sqrt {2} . Hence, the answer is boxed{2 sqrt {2}} . First, simplify each radical to its simplest form, then combine them. This problem tests the mixed operations of radicals: first, simplify each radical to its simplest form, then perform multiplication and division operations of radicals, and finally combine them. In the mixed operations of radicals, if you can combine the characteristics of the problem, flexibly use the properties of radicals, and choose the appropriate solution path, you can often achieve twice the result with half the effort.